Question

(TRIGONOMETRIA) Me ajudem???!! Expliquem passo a passo pfvr.

Answer 1

 Olá,
considere o segmento DB = y.  
 
 Do triângulo CBD:

$\tan60^o=\frac{x}{y}\Rightarrow\sqrt{3}=\frac{x}{y}\Rightarrow\boxed{y\sqrt{3}=x}$


Do triângulo ABC:

$\tan30^o=\frac{x}{100+y}\Rightarrow\frac{\sqrt{3}}{3}=\frac{x}{100+y}\Rightarrow3x=100\sqrt{3}+y\sqrt{3}\Rightarrow\boxed{y\sqrt{3}=3x-100\sqrt{3}}$


 Igualando-as,

$x=3x-100\sqrt{3}\\\\-2x=-100\sqrt{3}\\\\x=\frac{-100\sqrt{3}}{-2}\\\\\boxed{\boxed{x=50\sqrt{3}}}$

 Espero ter ajudado!

Answer 2

Seja $DB=k$

Triângulo DBC

$\text{tg}~60^{\circ}=\dfrac{BC}{BD}$

$\sqrt{3}=\dfrac{x}{k}$

$x=k\sqrt{3}$

Triângulo ABC

$\text{tg}~30^{\circ}=\dfrac{BC}{AB}$

$\dfrac{\sqrt{3}}{3}=\dfrac{x}{100+k}$

$3x=100\sqrt{3}+k\sqrt{3}$

Lembrando que, $x=k\sqrt{3}$:

$3x=100\sqrt{3}+x$

$2x=100\sqrt{3}$

$x=50\sqrt{3}$